This very basic idea of calculus requires a few lemmas before we are able to prove it.
If you are not familiar with the concept of a function, continuous function, or a closed interval, start here.
A function is said to be bounded if there is a value L such that for all x ∈ [a,b], f(x) ≤ L.
Lemma 1: Nested Interval Property for the real numbers
Suppose that I1, I2, ..., In is a sequence of nested, closed intervals where:
(a) Each Ii+1 is contained with Ii
(b) Each Ii interval is of the form [ai,bi]
(c) The lim (i → inf) (bi - ai) = 0 (See here for review of lim notation and the concept of limit)
Then:
There exists 1 and only 1 point c such that {c} = I1 ∩ I2 ∩ ... ∩ In
Proof:
(1) We know that there can be at most 1 number since lim(i → inf) (bi - ai) = 0 since:
If there were more than 1 number, the lim(i → inf) (bi - ai) would be greater than 0.
The only way that it can be 0 is if bi = c, ai = c, and c-c=0.
(2) We know that there is at least 1 number that is common to all intervals since:
(a) ai has a limit an that fits somewhere in the interval. For all intervals, it is clear that ai ≤ an ≤ bi
(b) Likewise, bi has a limit bn such that ai ≤ bn ≤ bi
(c) So we see that an,bn are both elements of all intervals.
(d) We further note that an = bn since the lim(i → inf)(bi - ai)= 0.
(e) So, if we let c = an = bn, then we are done.
QED
Lemma 2: If a function f is continuous on a closed interval [a,b], then f is bounded there.
(1) Assume that a function f is not bounded on [a,b]
(2) We can bisect the interval [a,b] into two halves which I will label I1 and I2.
(3) We can now pick an interval I which is unbounded. If both intervals are unbounded, then we can pick either one.
(4) We can repeat this process and create a sequence of nested, closed intervals which we can call Ii where each interval selects a subset which is unbounded.
(5) From Lemma 1, we know that there exists a point c which is common to all the intervals in #4.
(6) Because f is continuous, we know that there is a number ε such that f is bounded on the interval c - ε and c + ε [See here for definition of Continuous Functions]
(7) But one of the unbounded values in In must lie within (c - ε, c + ε)
(8) And this is a contradiction since from (#6), it must be bounded.
(9) Therefore, we reject our assumption.
QED
Lemma 3: Maximum value property of continuous functions
If a function f is continuous on the closed interval [a,b], then there exists a number c in [a,b] such that f(x) ≤ f(c) for all x in [a,b]
(1) Let I be the the closed interval [a,b]
(2) From Lemma 2, we know that I is bounded.
(3) Let λ be its least upper bound.
(4) We can divide I in half.
(5) At least one of these halves will have a least upper bound = λ (although it is possible that both have this least upper bound). Let I1 be the division of I which contains λ as the least upper bound.
(6) We can keep dividing up I1 in the same way until we have In which has λ as its upper bound and bn - an = 0.
(7) From Lemma 1, we know that there exists a point c which is common to all these intervals.
(8) It follows from (6) that f(c)=λ since:
(a) There exists a positive value δ such that if x-c is in between -δ and +δ, then f(x)-f(c) is between -ε and ε. [From the definition of a continuous function, see here]
(b) From (a), we have that f(c) - ε is less than f(x) which is less than f(c) + ε
(c) Since ε can be arbitrarily small, we can have f(c) ≤ f(x) ≤ f(c) which means that f(x)=f(c) at some point.
(d) In this case, f(c) cannot be more than λ since λ is an upper bound. [See here for the definition of an upper bound]
(e) Likewise, f(c) cannot be less than λ since λ is the least upper bound. [See here for the definition of a least upper bound]
(f) Therefore, f(c) = λ
QED
Definition 1: Right Hand Limit: lim (x → a+) f(x)
lim (x → a+) f(x) = L if and only if:
if x is between a and a + δ, then f(x) - f(a) is between -ε and ε
Definition 2: Left Hand Limit: lim(x → a-) f(x)
lim(x → a-) f(x) = L if and only if:
if x is between a - δ and a, then f(x) - f(a) is between -ε and ε
Lemma 4: One-sided and two-sided limits
The limit lim (x → a) for f(x) exists and is equal to the number L if and only if the one-sided liimits lim (x → a+) f(x) and lim (x → a-) f(x) both exist and both are equal to the number L.
Proof:
(1) Assume lim(x → a) f(x) = L
(2) Then if x - a is between δ and - δ, then f(x) - f(a) is between -ε and +ε [ By the definition of continuous functions, see here]
(3) Now x - a is between δ and -δ implies that:
x is between a - δ and a + δ.
(4) Since a is greater than a - δ (since δ is a positive value), this implies that:
if x is between a and a + δ, then x is necessarily between a - δ and a + δ.
(5) (#4) combined with (#2) gives us:
lim (x → a+) f(x) = L. [See definition 1 above]
(6) Since a is less than a + δ, this implies that:
if x is between a - δ and a, then x is between a - δ and a + δ.
(7) (#6) combined with (#2) gives us:
lim (x → a-) f(x)= L. [See definition 2 above]
(8) So, that proves the first half of the above theorem.
(9) Assume that lim (x → a+) f(x) = L and lim(x → a-) f(x) = L
(10) From definition 1, we have if x is between a and a + δ, then f(x) - f(a) is between -ε and ε
(11) From definition 2, we have if x is between a-δ and a, then f(x) - f(a) is between -ε and ε
(12) Combining (10) and (11) gives us:
if x is between a-δ and a+δ then f(x) - f(a) is between -ε and ε [For either condition (10) applies or condition (11) applies]
(13) Then, applying the definition of limit (see here), we get:
lim(x → a) f(x) = L
QED
Clarification: Local Maxima and Minima
A maximum or minimum is local if it is true for a given interval. A maximum or minimum is absolute if it is true across a domain. A maximum is value that is greater than all other points in for the range in question. A minimum is a value that is smaller or equal to all other points in the range in question.
Theorem: Local Maxima
If a function f(x) is differentiable at c and is defined as an open interval containing c and if f(c) is a local maximum value of f(x), then f'(c)=0.
Proof:
(1) Assume that f(c) is a local maximum value for f(x) on the open interval (a,b).
(2) Since c is differentiable, it means that right-hand and left-hand limits both exist and are equal to f'(c). [See Lemma 4 above]
lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)
lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)
(3) If Δx is greater than 0, then:
[f(c + Δx) - f(c)]/Δ x ≤ 0
This is true since f(c) ≥ f(c +Δx) for all small positive values of Δx since f(c) is a local maximum by assumption.
(4) If Δx is less than 0, then:
[f(c + Δx) - f(c)]/Δ x ≥ 0
This is true since in this case, we have a negative value f(c + Δx) - f(c) over another negative value Δx.
(5) So, we have from (#2):
f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≤ 0
(6) But we also have from (#2):
f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≥ 0
(7) Combining (#5) and (#6), we can conclude that f'(c) = 0 [since f'(c) ≥ 0 and f'(c) ≤ 0 ]
QED
Corollary: Local Minima
If a function f(x) is differentiable at c and is defined as an open interval containing c and if f(c) is a local minimum value of f(x), then f'(c)=0.
Proof:
(1) Assume that f(c) is a local minimum value for f(x) on the open interval (a,b).
(2) Since c is differentiable, it means that right-hand and left-hand limits both exist and are equal to f'(c). [See Lemma 4 above]
lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)
lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)
(3) If Δx is greater than 0, then:
[f(c + Δx) - f(c)]/Δ x ≥ 0
This is true since f(c) ≤ f(c +Δx) for all small positive values of Δx since f(c) is a local minimum by assumption.
(4) If Δx is less than 0, then:
[f(c + Δx) - f(c)]/Δ x ≤ 0
This is true since in this case, we have a positive value f(c + Δx) - f(c) over a negative value Δ x.
(5) So, we have from (#2):
f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≥ 0
(6) But we also have from (#2):
f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≤ 0
(7) Combining (#5) and (#6), we can conclude that f'(c) = 0 [since f'(c) ≥ 0 and f'(c) ≤ 0 ]
QED
References
- Edwards & Penny, Calculus and Analytic Geometry
No comments :
Post a Comment